Question

# PLEASE DON'T RESPOND IF YOU DON'T KNOW IT Consider the reaction 3Fe2O3(s) + H2(g)-----------2Fe3O4(s) + H2O(g)...

PLEASE DON'T RESPOND IF YOU DON'T KNOW IT

Consider the reaction

3Fe2O3(s) + H2(g)-----------2Fe3O4(s) + H2O(g)

Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.89 moles of Fe2O3(s) react at standard conditions.
S°surroundings =______ J/K

Given reaction is

3Fe2O3(s) + H2(g)-----------> 2Fe3O4(s) + H2O(g)

∆S°rxn=n x sum of ∆S° products - n x sum of ∆S° reactants

∆S°rxn=[2x∆S°Fe3O4(s) + ∆S°H2O(g)] - [3x∆S°Fe2O3(s) + ∆S°H2(g)]

∆S°rxn=[(2x146.44)+(188.72)] - [(3x87.40)+(130.59)] J/K

∆S°rxn=(481.6 - 392.79) J/K=88.81J/K.

For 3 moles of Fe2O3 react, ∆S°=88.81 J/K,

then for 1.89 moles Fe2O3 react, ∆S°=(1.89 mol x 88.81 J/K)/(3 mol)=55.95 J/K.

Please let me know if you have any doubt. Thanks.

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